Mathematics, 11.06.2020 20:57, newben

A company produces product with a mean weight of 10 and a standard deviation of 0.200. A new process supposedly will produce products with the same mean and a smaller standard deviation. A sample of 20 products produced by the new method has a sample standard deviation of 0.126. At a significance level of 10%, is it appropriate to conclude that the new process is less variable than the old?

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answered: CrownedQueen

$F=\frac{s^2_1}{s^2_2}=\frac{0.2^2}{0.126^2}=2.520$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_1 -1 =10-1=9 and for the denominator we have n_2 -1 =20-1=19 and the F statistic have 9 degrees of freedom for the numerator and 19 for the denominator. And the P value is given by:

Now we can calculate the p value with this probability:

$p_v =P(F_{9,19}2.520)=0.043$

Using a significance level of 5% we see that the p value is lower than this value and we have enough evidence to reject the null hypothesis and we can conclude that the variation for the new process is lower than the new one.

Step-by-step explanation:

Information given

n_1 = 10 represent the sampe size old

n_2 =20 represent the sample size new

s_1 = 0.2 represent the sample deviation for old

s_2 = 0.126 represent the sample deviation for new

The statistic is given by:

$F=\frac{s^2_1}{s^2_2}$

Hypothesis to test

We want to test if the new process is less variable than the old, so the system of hypothesis are:

H0: $\sigma^2_1 \leq \sigma^2_2$

H1: $\sigma^2_1 \sigma^2_2$

The statistic is given by:

$F=\frac{s^2_1}{s^2_2}=\frac{0.2^2}{0.126^2}=2.520$

Now we can calculate the p value with this probability:

$p_v =P(F_{9,19}2.520)=0.043$

answered: Guest

The answer is 41.7 because 1.39 times 30 is 41.7

answered: Guest

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05